题意:
机场快线有经济线和商业线,现在分别给出经济线和商业线的的路线,现在只能坐一站商业线,其他坐经济线,问从起点到终点的最短用时是多少,还有路线是怎样的;
思路:
预处理出起点到所有站的最短距离和终点到所有站的最短距离,枚举要坐的那趟商业线,然后里面最小的就是答案了;
AC代码:
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <bits/stdc++.h> #include <stack> using namespace std; #define For(i,j,n) for(int i=j;i<=n;i++) #define mst(ss,b) memset(ss,b,sizeof(ss)); typedef long long LL; template<class T> void read(T&num) { char CH; bool F=false; for(CH=getchar();CH<‘0‘||CH>‘9‘;F= CH==‘-‘,CH=getchar()); for(num=0;CH>=‘0‘&&CH<=‘9‘;num=num*10+CH-‘0‘,CH=getchar()); F && (num=-num); } int stk[70], tp; template<class T> inline void print(T p) { if(!p) { puts("0"); return; } while(p) stk[++ tp] = p%10, p/=10; while(tp) putchar(stk[tp--] + ‘0‘); putchar(‘\n‘); } const LL mod=1e9+7; const double PI=acos(-1.0); const int inf=1e9; const int N=2e5+10; const int maxn=500+10; const double eps=1e-6; int n,s,e,m; int mp[maxn][maxn],dis[2][maxn],vis[maxn],p[2][maxn]; inline void Dijkstra(int from,int flag) { mst(vis,0); vis[from]=1; For(i,1,n)dis[flag][i]=mp[from][i],p[flag][i]=from; dis[flag][from]=0; For(i,2,n) { int temp=inf,k; For(j,1,n) if(!vis[j]&&temp>dis[flag][j])k=j,temp=dis[flag][j]; if(temp==inf)break; vis[k]=1; For(j,1,n) { if(!vis[j]&&dis[flag][j]>dis[flag][k]+mp[k][j]) dis[flag][j]=dis[flag][k]+mp[k][j],p[flag][j]=k; } } } int main() { int Case=0; while(cin>>n>>s>>e) { if(Case)printf("\n"); Case++; For(i,1,n)For(j,1,n)mp[i][j]=inf; read(m); int u,v,w; For(i,1,m) { read(u);read(v);read(w); mp[u][v]=mp[v][u]=min(mp[u][v],w); } Dijkstra(s,0); Dijkstra(e,1); int ans=dis[0][e],uppos=-1,downpos=-1; int q; read(q); For(i,1,q) { read(u);read(v);read(w); int temp=dis[0][u]+w+dis[1][v]; // cout<<dis[0][u]<<" "<<dis[1][v]<<endl; if(temp<ans)ans=temp,uppos=u,downpos=v; temp=dis[0][v]+w+dis[1][u]; if(temp<ans)ans=temp,uppos=v,downpos=u; } if(uppos==-1) { int cur=s; while(cur!=e)printf("%d ",cur),cur=p[1][cur]; printf("%d\nTicket Not Used\n%d\n",e,ans); } else { int cur=uppos,cnt=0,tion[maxn]; while(cur!=s) { tion[++cnt]=cur; cur=p[0][cur]; } tion[++cnt]=s; for(int i=cnt;i>0;i--)printf("%d ",tion[i]); cur=downpos; while(cur!=e)printf("%d ",cur),cur=p[1][cur]; printf("%d\n%d\n%d\n",e,uppos,ans); } //printf("\n"); } return 0; }
原文:http://www.cnblogs.com/zhangchengc919/p/5689519.html